help-27

[untold bough]

wow i lost my stylys

okay i'm doing it tediously (digitally) then

wow it's a really nice solution

||8cos^4(x)||

[dusty mesa]

Well idk what i did

[untold bough]

what did you do

[untold bough]

yeah that just reduces back to your original eqn

doesnt work

[dusty mesa]

Yeah

Gotta simplify it more

[untold bough]

let's focus on one formula

since we're dealing with cosines

let's deal with cosines

cos(2x) = 2cos^2(x) - 1

well for cos(4x) just sub every x for 2x

cos(4x) is therefore 2cos^2(2x) - 1

[dusty mesa]

Just second

[dusty mesa]

How is this possible

[untold bough]

cos4x = cos(2(2x))

agreed?

[dusty mesa]

Okay

[untold bough]

let's stop confusing ourselves with x and 2x

cos2y = cosycosy - sinysiny

let y = 2x in this case

[dusty mesa]

Okay

[untold bough]

it's the same thing

[dusty mesa]

I technically did that

[untold bough]

that's a good first step

[dusty mesa]

Here

Second line

[dusty mesa]

Okay how do i continue ? Do i just use formulas more or?

[untold bough]

express everything in cos(2x) and simplify would be my next move

[dusty mesa]

Just second

Like this?

[dusty mesa]

Have you meant thay

[untold bough]

careless mistake

check your expansion

[dusty mesa]

Expansion...wait just second

[dusty mesa]

Ok sorry, does this look better

[untold bough]

expressing everything in terms of cos(2x) is simpler but you're on the right track

but your method will be more tedious

[dusty mesa]

So i guess...cos^2 (2x)= cos^4(x)-sin^4(x)

... right?

[untold bough]

it still works

[dusty mesa]

I don't know...how to. I am so dumb

[untold bough]

but now your powers are different

[dusty mesa]

So i can continue like that?

[untold bough]

yes

but it's unadvisable

[dusty mesa]

I just realised that

[untold bough]

okay i found a really crappy article

i'm gonna go but you can take a look at it

https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.800689.html

[dusty mesa]

Okay thanks

[untold bough]

same exact question

np

[dusty mesa]

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[steady onyx]

How can I find the oblique asymptotes of this function?

[gaunt galleon]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[gaunt galleon]

what exactly do you mean by oblique asymptote? A straight line?

[steady onyx]

In the question it tells me that there are multiple oblique asymptotes, my task is to find them

i have no more information then the function

and an oblique asymptote is like an asymptote that has a cartesian description

for example x = y

[acoustic leaf]

i would not expect that function to have an oblique asymptote

could you post a picture of the original problem statement?

[steady onyx]

okay just a second

[sand dove]

only vertical asymptotes

[sand dove]

yeah the drawing is messed up

[steady onyx]

so my teacher made a mistake

?

[scarlet sequoia]

there are formulas for the oblique asymptotes, are you familiar with them?

[steady onyx]

yes

[sand dove]

this is the most likely mistake

[steady onyx]

f(x)/x for a

and f(x)-ax+b to find b

[scarlet sequoia]

so to determine whether there are any or not you can apply them

[steady onyx]

starting with a?

[scarlet sequoia]

yes

[zenith spoke]

this is the correct curve

[steady onyx]

hmm ok

well i would still like to attempt because i wont have access to geogebra during my exam

[scarlet sequoia]

if e.g. limit doesn't turn out finite then an oblique asymptote won't exist

[steady onyx]

by finite you mean a tangible expression, like for example y = x?

so if i follow (sqrt(x^4-16)/2)/x I shouldnt get anything tangible?

[scarlet sequoia]

by finite I mean a number

like a = 2, 3, 1000, -5, sqrt(5) etc.

[steady onyx]

okay

[steady onyx]

because it seems like this should also have an oblique asymptote

[scarlet sequoia]

but this is actually doable in memory

but take your time and try to solve it then tell me your result

[steady onyx]

okay

well for a I get -1/2

[scarlet sequoia]

Show your work if possible.

[steady onyx]

okay i'll try and write it down nice, should pop up in a few mins

[scarlet sequoia]

I'm asking because it doesn't seem to be right

[steady onyx]

troubling

maybe I'm just silly

[acoustic leaf]

"infinity/infinity" is an indeterminate form, so you can't just say it evaluates to 1 out of hand

[steady onyx]

oh

what can i do then?

[acoustic leaf]

consider that x^2/x = x for all x ≠ 0

[steady onyx]

well then i would have x*sqrt(1-16/x^4)

and I dont exactly know what to do with that

[acoustic leaf]

take the limit as x approaches infinity

[steady onyx]

well then you would get infinity * 1

aka infinity

[acoustic leaf]

so what can we say about our oblique asymptote if the limit implies it has infinite slope?

[steady onyx]

that its a vertical asymptote?

i dont really know

because infinity times a always equels infinity

[acoustic leaf]

well we assumed that there must be a diagonal line that the function approaches, but the slope of such a line is not defined, so can such a line exist?

[steady onyx]

no

[acoustic leaf]

so the function does not have an oblique asymptote

[steady onyx]

oh i follow

interesting

[steady onyx]

then what kind of asymptotes would it have then?

i cant see a horizontal one

nor any vertical ones

although i suppose the whole area of ]-2,2[ is kind of an asymptote

[acoustic leaf]

i would conclude that it has none

[steady onyx]

okay

thank you for your help, and everyone else beforehand. I appreciate it a lot!

.close

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[vestal lance]

How to do the last step?

[vestal lance]

My guess is that they’re using the thin svd somehow but I’m not too sure how they got the matrix

[elfin hill]

What's $\Sigma_{r}$ in that context?

[woven radish]

cjg#1618

[vestal lance]

It is the square diagonal matrix containing all singular values in \Sigma

[devout snow]

@vestal lance Has your question been resolved?

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[kindred talon]

How do i do 6? I just want a starting guide not the full answer but i dont know where to start

[acoustic helm]

Do you know what the inverse of log is?

[kindred talon]

Ln

Right?

[nova glen]

no

[acoustic helm]

ln is a log

[kindred talon]

Waut expo ent

Exponent

So 1 to the power of each?

[nova glen]

the base of just log depends on convention. i just assumed it to be e

[acoustic helm]

Most likely base 10

[kindred talon]

Base 10 log

[acoustic helm]

Then what would be the inverse?

[kindred talon]

10^7x+1??

[acoustic helm]

Well you would have 10^(log(…))

[kindred talon]

I'm confused

[acoustic helm]

If you have log(x) then you can apply base 10 so you have 10^log(x). Then the base 10 cancels the log

[restive river]

You can undo the logarithmic functions by setting both as the power of base 10

[frosty cradle]

but you need to change the 1 to log of something (well, you don't need to I guess)

[kindred talon]

This??

[frosty cradle]

no, 10 to the power of the entire RHS

[kindred talon]

The +1 is excluded from the paraenthesis

Fixed it

[frosty cradle]

you still need the logs there, too

[kindred talon]

Oh

So what would i do though?

10^1-1?

[acoustic helm]

[ \log (\dots) = \dots ]
Apply base 10
[ 10^{\log (\dots)} = 10^{\dots} ]

[woven radish]

shsgd

[acoustic helm]

Then the 10 cancels the log, which leaves just the dots

[kindred talon]

I need to see the dots as x gimmie a sec

[ \log (x) = x ]
Apply base 10
[ 10^{\log (x)} = 10^{x} ]

[woven radish]

Sprout

[kindred talon]

Yeah but what about the leftover 1?

[acoustic helm]

you are applying base 10 to the entire equation so you have 10^{log(x) + 1}

what power law can you apply to deal with the 1?

[kindred talon]

Zero?

10^1-1

Or is that wrong

[acoustic helm]

Recall ( x^{a+b} = x^a \cdot x^b )

[woven radish]

shsgd

[kindred talon]

,rotate

[kindred talon]

This? Ignore bottom 10

[acoustic helm]

well remember you are applying base 10 to the entire equation, so you include the logs

,, 10^{\log (7x+1)} = 10^{\log (x-2) +1}

[woven radish]

shsgd

[kindred talon]

Uhuh?

Ive written that down so now do i cancel the log out? To get 10^ whatv?

[acoustic helm]

,, a^{\log_a (x)} = x

[woven radish]

shsgd

[acoustic helm]

use this ^

[kindred talon]

,rotate

[acoustic helm]

second last line can be crossed out

and remember what you have to do with the + 1 before cancelling the logs

[kindred talon]

Make it a 10?

[acoustic helm]

👍

[frosty cradle]

it's still not right

[kindred talon]

Yeah

Its supposed to be 7

I plugged it into a math solver

But idk why 😭

[acoustic helm]

after using the exponent law on the + 1 what do you have?

[kindred talon]

10

[acoustic helm]

I mean what do you get for the entire equation

[acoustic helm]

not + 10, remember this is the rule

[kindred talon]

I redid it

[acoustic helm]

Yeah that's definitely one way to do it

[kindred talon]

So i see what i went wrong

The two should have been a 20

And the x should be 10x

Actually nvm

But idk

[acoustic helm]

,, 10^{\log (x-2) + 1} = 10^1 \cdot 10^{\log (x-2)}

[woven radish]

shsgd

[kindred talon]

Yeah

I didnt understand that just then but now it makes sense

[acoustic helm]

then cancel the log and you have
10(x-2)

[kindred talon]

Yeah

I'm slow 😔

Sorry for taking up your time

Thank you though!! 😊💌

[acoustic helm]

I suggest you try more problems that use power laws. Getting comfortable with them will really boost your algebra skills

[kindred talon]

Tyty

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[lusty fox]

So lim h->0 f(x+h)-f(x)/h means there’s a small arbitrary change at h but why do we say h->0 ? Like what about if we say the change is about h=0.5 !!

[lusty fox]

I know it’s a small arbitrary value near to 0 because it have to be a linear line

But why do we say 0?

why do we evaluate h at 0

[graceful cosmos]

We are exactly concerned with changes that are "as small as possible"

[graceful cosmos]

Note we don't evaluate at h = 0. We take the limit as h approaches 0. So "a very small h".

[lusty fox]

Oh so we approximate it ?

like from the biggest numbers

[graceful cosmos]

Approximate which, sorry?

[lusty fox]

the h

If we don’t evaluate it at h=0 then why do we take this limit ?

And how is it still possible to give us the difference ? The secant line would start at 0 if we take the limit (evaluate) it at h=0

[outer kestrel]

we cant evaluate at 0

[lusty fox]

Oh

yes because it’s improper function

It will diverge

[graceful cosmos]

You might be trying to equate "evaluate at 0" and "take the limit to 0". They are not the same thing.

Specifically, evaluation fails. You can't plug in h = 0 without a division by 0.

But the limit can succeed.

[lusty fox]

What’s the correct definition of Take the limit then

[outer kestrel]

search the epsilon delta definition

its the formal definition

[graceful cosmos]

There's a specific definition called the "Epsilon-delta definition". If your class hasn't shown it to you, you likely don't want to see it

[outer kestrel]

intuitively, as h gets closer snd closer to that value

[outer kestrel]

ye i dont think elementary calc curriculum teaches epsilon delta

[graceful cosmos]

But you can think of the limit as "letting h get really small, and seeing where (f(x+h) - f(x)) / h goes"

[lusty fox]

Oh , alright so suppose we say h=0.5 and f(x) is defined at x=1 and then the rate of change is going to be x=1.5 and the secant line will start from x=0.5 and end at 1.5 right ?

[graceful cosmos]

If you plug in x = 1 and h = 0.5, you'll find the slope of the secant line between x = 1 and x = 1.5

You're basically using the slope formula

[devout snow]

@lusty fox Has your question been resolved?

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[proven geyser]

Is this a wrong way to go about the problem or is this ok

[mystic sluice]

e^(lna)=a

[hexed summit]

You can cancel out the ln by raising e

[mystic sluice]

i think

[hexed summit]

yeah

e^(ln(3)) = e^(ln(x))

would be the appropriate step to write down

[proven geyser]

Ah ok

[mystic sluice]

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[proven geyser]

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[red shale]

how do you get the second equation

i mean x value i

i understand that first is 42 degrees

but like

i thought you just add 90

[devout snow]

Show your work, and if possible, explain where you are stuck.

[mystic sluice]

factorize it

(sin x-3)(3sin x-2)

=0

so sin x=3 / sin x=2/3

and sin x != 3

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[mystic sluice]

dude its !=

nvm why am i arguing with a bot

[devout snow]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[mystic sluice]

!learn

!help

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[mystic sluice]

!answer

!process

[devout snow]

@red shale Has your question been resolved?

[devout snow]

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[restive river]

can somone explain what rafilou tried to say bc in can understand nothing

[orchid sierra]

@restive river the secret has to do with the properties described after each of these definitions:
https://en.m.wikipedia.org/wiki/Bijection,_injection_and_surjection

[restive river]

hi melvin

i actually know the properties of these of kind of functions

my school didnt really tell me about the properties tho but i know the definitions

but let me go through it

[orchid sierra]

I never understood these things, I was always a little off for at least one of these

[near jolt]

rafilou plugged in x=f(y)

f(f(y)-f(y))=f(f(y))+f(y)^2+f(f(y))-1

rearranges into 2f(f(y))=-f(y)^2+f(0)+1

[restive river]

ok i get what you mean qwerty

but can you tell me how they got this

bc theyre correct, the ans is 2

is that some kind of propery melvin just talked about

[near jolt]

if f is surjective then there exists some value c such that f(c)=0

then we plug in y=c and get
2f(f(c))=f(c)^2+f(0)+1
since f(c)=0 this gives 2f(0)=f(0)+1 which means f(0)=1

[near jolt]

now plugging f(0)=1 into 2f(f(y))=-f(y)^2+f(0)+1 we get 2f(f(y))=-f(y)^2+2

[restive river]

but man how do we know that f is surjective

is there any other way?

[near jolt]

Furthermore due to surjectivity for any x there exists y such that f(y)=x and plugging x=f(y) gives you the desired result

[restive river]

explain this a bit more pls

[restive river]

im clear till this step

[near jolt]

Since f is surjective, for every u in R there exists a v such that f(v)=u

[restive river]

true

[near jolt]

plugging in y=v we get
2f(f(v))=-f(v)^2+2
but we have f(v)=u
so 2f(u)=-u^2+2

since u is arbitrary, f(u)=(-u^2+2)/2 must hold for all real number u

which means that the function must be f(x)=-x^2/2+1

@restive river is this clearer?

[restive river]

hold on lemme process a bit

@near jolt im able to understand everything

i have a one more question

from this question itself

[restive river]

the only 'if' we are using here is that f should be surjective, and if its surjective this is true

[near jolt]

yeah, we have not proved that it is surjective (i dont even know if it is necessarily surjective)

[restive river]

or are there other thing too apart from f(c)=0?

man but what if my teacher asks how i know its a surjective or not

out answer is totally based on an assumption

[near jolt]

im aware

there is probably a better method (or a way to prove it is surjective)

[restive river]

yea true

still, thanks for the help

[near jolt]

ill dm u if i think of something ig

[restive river]

ill also ask my teacher about this question, im curious to see what's the other method

[restive river]

ok i have class today, ill tell you if my teacher tells me some other method

alr can i close the channel now?

[near jolt]

yes

[restive river]

ok

.close

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[solid osprey]

f(x) is a polynomial with non negative real coefficients, if f(6)=24 and f(24)=1536, what is the maximum value of f(12)?

[solid osprey]

this might(?) be coorelated with cauchy-swarzh inequality since its the topic, but it can sometimes stray way from it

[devout snow]

@solid osprey Has your question been resolved?

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[golden drum]

hi guys any idea ofr ii)

[golden drum]

i have no clue how to approach this

[sharp harness]

which one?

[golden drum]

ii)

please

[sharp harness]

these are vectors right?

you can sum up everything

n-1 to 1 - k-1 to 1

wait

are these vectors?

[safe jasper]

these are binomial coefficients

[sharp harness]

i don't think so

[sharp harness]

yeah

otherwise it won't make any sense

let me think wait a min

[golden drum]

teah

they are

its all binomial stuff this worksheet

[sharp harness]

I think I got it

[golden drum]

rly

okok

[sharp harness]

you should use the sum property

i think it is

[golden drum]

um

[sharp harness]

nCr + nCr-1 = n+1 C r

[golden drum]

mhm

yes i know that one

[sharp harness]

can you check this property pls if it is correct?

[golden drum]

like

if thats the right thing

?

[sharp harness]

yeah

[golden drum]

okok

wai

the property says

ncr = n-1Cr-1 + n-1Cr

so

same thing

yeah

[sharp harness]

ok

now see the last term

[golden drum]

oh yes

thats equal to just 1

[sharp harness]

change k-1Ck-1 to kCk

[golden drum]

oh

[sharp harness]

kCk-1 + kCk = k+1Ck

similarly keep adding

you will reach

n-1Ck-1 + n-1Ck = nCk

I think it solves the problem

[golden drum]

hm

okok

i will have a look

but thank u

.close

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[native heart]

can some1 explain this to me? im a little confused since i guess why does this imply that the non-diagonal entries are 0

[native heart]

i guess i was thinking like if A = 3x3 with nondiagonal entries nonzero and the diagonal entries are B1 B2 B3, i dont see why the hypothesis holding implies A diagonalizable

[devout snow]

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[crude solstice]

Let A and B be intersection points on the line y=2x+1 and the circle (x+1)^2 + (y-2)^2 = 5. The abscissa of point A is less than the abscissa of point B. Find a point C on the circle such that angle ABC is a right angle

[crude solstice]

$y = 2x+1 \ (x+1)^2 + (y-2)^2 = 5$

[woven radish]

nashira._.

[crude solstice]

by substituting y in:

$(x+1)^2 + (2x+1-2)^2 = 5 \ 5x^2 -2x -3 = 0$

[woven radish]

nashira._.

[crude solstice]

$x = 1, y = 3 \ \ x = -\frac{3}{5}, y = -\frac{1}{5}$

[woven radish]

nashira._.

[crude solstice]

so since the x value of A will be less than B it means the top one is point B and the bottom one point A

so

$A \ (-\frac 35, -\frac 15) \ \ B (1,3)$

[woven radish]

nashira._.

[crude solstice]

so now point C lies on the circle in such a way that angle ABC is a right angle

so I guess

$B \perp A$

[woven radish]

nashira._.

[crude solstice]

so I guess

$y_A = m_Ax_A+c_A \ y_B = m_Bx_B+c_B$

since they are perpendicular

$m_A \times m_B = -1$

[woven radish]

nashira._.

nashira._.

[crude solstice]

yeah im kinda stuck here

I mean I know I could probably substitute into m_A or m_B but that is yielding no results

[devout snow]

@crude solstice Has your question been resolved?

[crude solstice]

<@&286206848099549185>

[devout snow]

@crude solstice Has your question been resolved?

[restive river]

Won't you get a relation here ?

[dense lynx]

you could use this relation to construct the line perpendicular to AB that passes through B (you know the slope and a point it passes though) and plug that line into the circle to find C

but that's not the best way; recall properties involving circles and right angles

[crude solstice]

i know that if there is a right angle it means that the side opposite to the right angle should be the circumference

[dense lynx]

the diameter

[crude solstice]

oh yeah the diameter lmao

So I'd guess C has to be connected with A in such a way that angle B is 90 degrees now that way I have no idea

the diameter would be

$d = 2r = 2\sqrt 5$

[woven radish]

nashira._.

[crude solstice]

and I guess it passes through the center (-1,2)

yeah i still have no idea

[devout snow]

@crude solstice Has your question been resolved?

[crude solstice]

in this case we know that the values of A and B are:

[crude solstice]

$A \ (-\frac 35, -\frac 15) \ \ B (1,3)$

[woven radish]

nashira._.

[crude solstice]

so since we know A and B how would I solve for C?

[devout snow]

@crude solstice Has your question been resolved?

[devout snow]

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[agile elk]

if I do a one-dimensional discrete cosine transform on a line of 8 values ​​and I do two dcts separately for the even and odd values, how do I put the transformed values ​​together?

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[kindred kernel]

So i was messing around with a specific problem i saw that asked you to count how many times you write down the number 1 between 0 and some other number, which for this question was 99,999. I managed to come up with a formula that roughly gives how many times you'd have to write the digit one (or any digit for that matter) before you get to that value, but it's not very accurate except for around the powers of 10. I wanted to ask two questions: is there a way I can use this to also get a rough value for how many numbers have the digit one in them (as in, 111, does not count as 3 repetitions), and is there a more accurate way to do this?

[kindred kernel]

$round(log(x) 10^{log(x)-1})$

[woven radish]

Serphic

[uncut crow]

can you give some more examples

can't tell what you're asking for

nvm i think i understand now

you can probably do this combinatorially

taking 99999 for example, we can consider 5 length strings of elements from {0,1,...,9} (leading zeros allowed)

it would be messier for other numbers tho. 99999 is nice

[uncut crow]

but anyway there are 10^5 of these, and 9^5 that have no 1s. so the answer would be 10^5 - 9^5 (maybe off by 1 or 2 depending on if 0 and 99999 are included in the count or not)

[kindred kernel]

i may be a little stupid, but would there not be 10^5 different elements rather than 5^10

[uncut crow]

oh oops

[polar chasm]

Also i dont think it works?

I managed to get recursive formula

ohh i see now

[uncut crow]

no i am the stupid one

[kindred kernel]

yeah i have a recursive formula but its.. really weird

its like an infinite product or something

[polar chasm]

^ this one here is good

In numbers from 0 to 10^n - 1, there are 10^n - 9^n numbers with a 1 in them

[kindred kernel]

w h a t

okay hold on

lemme process that

[polar chasm]

There are 10^n numbers between 0 and 10^n - 1

[kindred kernel]

mhm

i get that

[polar chasm]

now how many of them don't have 1s?

note that those are exactly the numbers formed using those digits: {0, 2, 3, 4, 5, 6, 7, 8, 9}

[kindred kernel]

wait why is that so easy 😭

no that makes sense

[polar chasm]

Did you get it?

[kindred kernel]

it feels a little too simple to be the solution but yeah i get it

wait so according to this formula, as the number of digits increase, most numbers start to have at least one of every digit in them, right?

[polar chasm]

Hmm why according to this formula?

I dont get the connection

[kindred kernel]

limit as x goes to infty of (10^x - 9^x)/10^x = 1

[polar chasm]

ah

well yeah

thats quite obvious fact

the more digits there are, the higher is the probability that there is at least one of every digit

[uncut crow]

if you pick a random 1000 digit number it's probably going to have every digit

[kindred kernel]

no but it is cool to have a (probably not extremely accurate) estimate of that probability

[polar chasm]

This counts just ones btw

[kindred kernel]

not necessarily just ones tho

we could do 2 for this and it should be the exact same formula, no?

[polar chasm]

Yeah

all digits except for 0 have the same formula

but you cant get the estimate of probability that there is at least one of every digit just with this formula

[kindred kernel]

valid
it does say that there are only 4 ones between 1 and 30 💀

[polar chasm]

Incorrect application

[kindred kernel]

yeah

im aware of that

okay so

[uncut crow]

this formula only works for powers of 10

[kindred kernel]

one last question, and im not sure if this is even valid

could i apply this in other bases by changing the base of the exponent and base of the log

so like

one sec

im just using log so that its numbers rather than powers of the base

would this hold true for like.. base 5 or something

yknow what actually nvm i think i have my answer lmfao

.close

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[restive river]

if theres an infinite amount of real numbers between any 2 real numbers, how does time pass? how would 1 second become 2?
wouldnt an infinite amount of time need to pass for that to happen?

[restive river]

im not very knowledgeable on this so pardon me if my question is absurd

[uncut crow]

evidently not

[restive river]

how about theoretically?

[magic thicket]

each instant has 0 length, so you've got infinitely many instants of length 0
It could be that this indefinite instance of "0 * inf" happens to be the length of the interval

[restive river]

Ohhh youre rightt, because every instant is a part of the other

[sharp harness]

why are you mixing time with numbers?

[restive river]

because we measure time with numbers ig

[sharp harness]

you can't say that, time can be quantized, we simply don't know yet

[restive river]

number of seconds, minutes, hours

[sharp harness]

please don't mix physics with mathematics

[sharp harness]

^

[restive river]

interesting

why do we assume it is tho?

the invention of the clock is based on that idea no?

[sharp harness]

also i would say don't mix physical quantities with mathematics, even length can be quantized (planck's length), we simply don't know

[restive river]

you mean cant be?

are there any examples of a physical measurement that can be quantized?

at all?

that we know of

[magic thicket]

that doesn't prevent physicists and basically everyone else from considerings things that move as a function of time, and have a defined position at every real number

[sharp harness]

charge

[restive river]

can charge go from 1 to 2?

ig

[magic thicket]

and even if so, that doesn't prevent us from thinking about the possibility of a real-indexed time

[sharp harness]

I mean you don't take charges based on the charge of quarks, do you?

[restive river]

this makes complete sense tho

[sharp harness]

when you have many, you can assume it continuous, but at small scales, they are not

[restive river]

intuitively at least

[sharp harness]

they are quantized

[magic thicket]

exactly, it's just a real number

99.99% of the time, we are totally free to ignore quantization

[sharp harness]

it depends, if you are a particle physicist, 0.01% of time you are free to ignore quantization

if you are just an engineer 99.99% of the time, you are totally free to ignore quantization

[magic thicket]

and particle physicists are that 0.01% of the people on earth

[restive river]

anywho, thanks a lot guys

I appreciate it

ill close this now

feel free to take it to #discussion

.close

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[opaque talon]

How do I do part a

[vestal dirge]

For independent events \( A \) and \( B \), the probability of their intersection is given by:

\[ P(A \cap B) = P(A) \times P(B) \]

[woven radish]

tobi

[devout snow]

@opaque talon Has your question been resolved?

[opaque talon]

but I did 0.1= (0.1p)(0.4) and i got it wrong

[unkempt warren]

0.4*(0.1+p)=0.1

0.1+p = 0.25

p = 0.15

[opaque talon]

oh fucking hell

I said 0.1 times 0.4 was 0.4 in my maths exam

😭

thanks

.close

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[urban wigeon]

Yo can I get help with this

[proud perch]

first of all you can factor out the 5

second of all you can bring the expression to a common denominator

then just use L'H

[urban wigeon]

Ah k thanks

[raven leaf]

how do you get l'h?

[urban wigeon]

L hospital

[raven leaf]

yeah how

[proud perch]

by bringing it to a common denominator

[raven leaf]

I'm not sure the conditions are satisfied

[proud perch]

ignoring the 5 you should get:
(1 - sqrt(1+t)) / (t sqrt(1+t))
which is a 0/0 limit

[urban wigeon]

I think he saying find the derivative

[raven leaf]

only one item tend to 0

[proud perch]

it's a 0/0 limit

but just remember to multiply by 5 afterwards

[raven leaf]

the top would be 1?

[proud perch]

1 - 1 = ?

[raven leaf]

oh my bad

i thought you meant bring out 5/t

gotcha

[urban wigeon]

Do u mutiple the 5 to both numerator and denominator

[raven leaf]

just take it out of the limit

[urban wigeon]

Ah I see bc the limit of itself it just itself

[raven leaf]

thr limit of 5 is 5 yes

[urban wigeon]

I got it’s -5/2

[devout snow]

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[quiet tide]

can someone breakdown the first step

[quiet tide]

-1 +1?

[dense jay]

adding 0 presumably to get everything into a more easily workable form

they probably split the fraction next

[quiet tide]

very confusing for me

[dense jay]

the limit to 0 for (a^x-1)/x is ln(a)

so they just got it into a form where you can apply that

to get ln(25)-ln(16)

[quiet tide]

ohhh

why did
ln25 - ln16
turn into
ln25/16?

[dense jay]

thats just basic properties of logs

if youre doing limits id presume you know the log laws at least

[quiet tide]

honestly dont

[dense jay]

oh

[quiet tide]

barely passing calc, i have exam on wednesday

[dense jay]

this essentially

number 5

[quiet tide]

oh thank you!

and then (5/4)^2 which i guess equals 25/16

[dense jay]

you guess?

my friend this is stuff you should be certain of

[quiet tide]

yeah iam super bad at math tho 🙁

aight thanks alot, that picture would also help a ton!

.close

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[willow sedge]

Can someone explain how 2.667 and 7 make 29/3?

[willow sedge]

When I plugged it into the calculator I got a huge fraction

[thin fern]

$2.\overline{6}$ (or 2.67 approximately) is $\frac{8}{3}$ as a fraction

[woven radish]

Ari

[thin fern]

$7 = \frac{21}{3}$

[woven radish]

Ari

[willow sedge]

But

[thin fern]

Yeah, because 2.67 is an approximation, not the actual decimal value

Desmos won't give you the exact value

The exact value is $2.6666...$

[woven radish]

Ari

[willow sedge]

Hmmm

But how am I supposed to know how many 6’s there are

[thin fern]

there are infinitely many

hence the ...

The idea is that its equivalent to 2 2/3, or 8/3

[willow sedge]

How do I get that

[thin fern]

Since $1 = 0.999...$, $\frac{1}{3} = 0.333...$ and so $\frac{2}{3} = 0.666...$

[woven radish]

Ari

[willow sedge]

Yeah but it’s not always a nice decimal

[thin fern]

sure but in this case it is

[willow sedge]

I see

What would I do if it’s not a nice decimal

[thin fern]

you probably won't be asked to convert those into fractions then

One not-as-nice decimal you should know is $\frac{1}{7} = \overline{0.142857}$

[woven radish]

Ari

[willow sedge]

I see

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[brisk urchin]

two lines r parralel
prove that rectangle (abef) same as square (acd 0,0) no matter the value of b

[devout snow]

@brisk urchin Has your question been resolved?

[brisk urchin]

<@&286206848099549185>

[thin inlet]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[brisk urchin]

1

[thin inlet]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[thin inlet]

i assume you mean value of y? not value of b?

[brisk urchin]

Nope

the question asked me to prove that the square always equalled to the triangle no matter the coordinates of B

[thin inlet]

equal area?

[brisk urchin]

yeah

What i know so far is that B is for sure ( x , 24)

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[devout snow]

@brisk urchin Has your question been resolved?

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[visual torrent]

hi this is a question with its answer, it asks to find if series converges or diverges

[visual torrent]

can someone explain where they got n^5/4 from?

why not just use 1 / n^3/2 ?

[safe jasper]

if they would have compared to 1/n^(3/2), then the limit would simplify to the limit of (ln(n))^2

which diverges

ln(n)/n^(3/2) is greater than 1/n^(3/2)

just because you know 1/n^(3/2) converges, this comparison is useless

all you know is that ln(n) / n^(3/2) is greater than a convergent series, that doesn't tell you whether or not it converges

but when you pick 1/n^a where a > 1 and a < 3/2 and you do the limit comparison test, you'll find that the limit goes to 0

they didn't have to use 1/n^(5/4)

but notice that if they used 1/n^a with a <= 1, it would diverge

so a > 1

and if a >= 3/2, then the limit would diverge

[visual torrent]

i see thanks

[visual torrent]

wait you mean converge right?

if a > 1 then it converges

[safe jasper]

1/n^a converges when a > 1, yes

i mean the limit used in the limit comparison

((ln(n))^2 / n^(3/2)) / (1/n^a)

= n^a * (ln(n))^2 / n^(3/2)

that diverges when a >= 3/2

which isn't useful for the comparison

since you want the limit to go to 0

[visual torrent]

got it

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[idle coral]

Hello, good morning, please help me to understand this resolution, it asks to determine for which values n (natural) the limit is finite?

[idle coral]

I don't understand the parts of the inequalities or what they were raised for by whoever did it

[tall stirrup]

Hello, good evening, does sen(x) mean sin(x)

[idle coral]

yes

spanish

hi

[tall stirrup]

so, the |x| >= |sin x|

we introduced this so that the expression simplifies

[idle coral]

hmm ok

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[idle coral]

.close

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[dry rover]

Hello again, we want to solve a² + b² = c² in $Z^3$. $A$ is the ring $\mathbb{Z}[i]$, $a$ is odd and $b$ is even, (a +ib) and (a -ib) are comprime, a and b they're both elements of $\mathbb{Z}$ and there exists $\epsilon \in U(A)$ and $\beta \in A$ such that $a + ib = \epsilon \beta^2$; why can't $\epsilon$ be $-i$ or $i$?

[woven radish]

abyssworld

[sand dove]

imaginary part of beta^2 is even

has to coincide with a being odd and b being even

[dry rover]

so it changes the parity of a and b

i and -i

[sand dove]

under the hypothesis that a is odd and b is even, you can't change their parity

[dry rover]

and you meant to say that if epsilon is i or -i, a should be even and b odd? how so?

or

[sand dove]

no, the opposite

if a is odd and b is even

you cannot have epsilon be -i or i

since the parity of a and b is fixed

[dry rover]

ok, so if epsilon = i; we have a + ib = -x² -y² - 2iy if beta = x + iy; so

the imaginary part is still even here

if epsilon = i

[sand dove]

no

you would have i(x^2 - y^2 + 2ixy)

so real part is -2xy

but a is odd

[dry rover]

yes

[sand dove]

impossible

[dry rover]

ok thank you 👍

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[old carbon]

Hi

[old carbon]

extremely basic question

sin^2 h/2

does that mean sin (h/2)^2 ?

[jolly socket]

what level is this?

[old carbon]

wdym?

[upbeat pawn]

(sin (h/2) )²

[old carbon]

how do i open this ?

that means

i mean sin h^2/4

right?

[nova glen]

no

[upbeat pawn]

Wha

[old carbon]

sin(h^2/4)

[nova glen]

it means sin(h/2)*sin(h/2)

[woven radish]

y0shi

[old carbon]

wow ofc it means that

bruv

[old carbon]

how do i solve it further without making it sin^2 h/2 again?

[nova glen]

half angle formulas

[old carbon]

would you please elaborate ?

look

i know it will open to something like sin h/4

i mean sin h^2/4

[nova glen]

$\sin(\frac{x}{2}) = \sqrt\frac{1-\cos(x)}{2}$

[woven radish]

esca (@ with reply)

[old carbon]

wait fuck yeah

[old carbon]

sir no

[nova glen]

fym no

[old carbon]

i don't want to fuck with cos rn

can't

[nova glen]

whats the full problem

[old carbon]

why did you say no to this?

[old carbon]

im trying to make a formula, there's no quesiton sir

[nova glen]

because sin^2(h/2) is not the same as sin(h^2/4)

[old carbon]

elaborate ?

[old carbon]

how old are you? wait

[old carbon]

are you sure about this? im positive that u are wrong here

[nova glen]

explain why

[old carbon]

because then i can make it 0 if h= 0 and with that ican move aheadd

[nova glen]

it is 0 when h = 0

[old carbon]

no no no

im talking about different h

no

dude

[nova glen]

i dont understand what youre trying to say

[old carbon]

wait

sin ^2 h/2

= sin(h^2/4)

right?

[nova glen]

no

[feral agate]

$\sin^2\left(\frac{h}{2}\right)=\left(\sin\left(\frac{h}{2}\right)\right)^2\ne\sin\left(\left(\frac{h}{2}\right)^2\right)$

[woven radish]

kheerii

[old carbon]

bruh?

what

[nova glen]

take notes

[feral agate]

You’re squaring the whole function, not the input

That’s how trigonometric notation works

[old carbon]

wait

let me give you a better one

sin^2 h/2
/
4 * h^2/4

solve it

[feral agate]

What..?

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[nova glen]

are you trying to find

$\sin^2(\frac{h}{2}) = h^2$
?

[woven radish]

esca (@ with reply)

[old carbon]

NO

bruv

[feral agate]

$\left(\right)$ for brackets around fractions

[woven radish]

kheerii

[nova glen]

ah ty

[old carbon]

NO

qIR

WAIT

[nova glen]

i dont understand what youre asking then

[old carbon]

okay

prove (sin^2h/2 ) / (4* h^2/4) = 0

[feral agate]

That’s not even true..

[stable storm]

why’s h^2 being multiplied and divided by 4..

that’s just h^2 at the bottom

[nova glen]

$\frac{\sin^2\left(\frac{h}{2}\right)}{h^2}=0$

[woven radish]

esca (@ with reply)

[stable storm]

doubt that’s a true identity

[old carbon]

then i can cut it from the upper half somehow

[nova glen]

its definitely not lol

this means sin^2 of everything is 0

[feral agate]

@old carbon

Don’t ask random questions in text form

[old carbon]

if h = 0

if i take this

[feral agate]

No

[old carbon]

then is it possible?

[nova glen]

no

then its undefined

[old carbon]

EXACTLY

[stable storm]

is the question asking for the limit as h approaches 0 of that?

[old carbon]

that's why i did multiplied and then divided by 4

[old carbon]

yeah kinda

[stable storm]

why’re you plugging in 0 randomly

[nova glen]

wdym kinda

[old carbon]

wait

let me do one better

[feral agate]

That’s a completely different question

[nova glen]

im guessing this is a language barrier

[stable storm]

probably

[old carbon]

chill guys, I know how to speak ingles, i just can't find the question right now

hold on

[feral agate]

$\lim_{h\to 0}\frac{\sin^2\left(\frac{h}{2}\right)}{h^2}$

Is this your question?

[woven radish]

kheerii

[old carbon]

that helps ?

this is how it is supposed to be

[nova glen]

okay what is q

[old carbon]

@stable storm @feral agate @nova glen

[nova glen]

is this the original problem?

[old carbon]

i don't have the question, but i can assure you the above thing is 101% correct

[old carbon]

i just don't know how we went from q/2 to q/8

[nova glen]

the limit is 1/4

[old carbon]

elaborate?

[nova glen]

well this expression is 1/4. you can find it with l'hopitals or squeeze theorem i think

[feral agate]

No need

[stable storm]

just separate it

[feral agate]

$\lim_{h\to 0}\frac{\sin h}{h}=1$

[old carbon]

how

[woven radish]

kheerii

[nova glen]

what kheeri said

its an identity

or uh

something that i had memorized

[old carbon]

i have it too

but still i don't know how we went to q/8

[stable storm]

$\frac14\lim_{h\to 0}\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$

[nova glen]

(q/2)(1/4) = q/8

[old carbon]

wait

yeah

FUCK

[old carbon]

right

[woven radish]

y0shi

[feral agate]

[old carbon]

hm

right

q/4 */14

yeah

[feral agate]

Next time just send us a picture of your question directly

[old carbon]

wtf

[feral agate]

Instead of trying to type it out and failing

[old carbon]

i was struggling with this from last 1 hour, it seems like i should sleep

thanks guys

it helps

[nova glen]

all this is assuming q isnt dependent on h. but yeah np

[old carbon]

yea it isn't

[nova glen]

alright

close the channel and go to sleep

[old carbon]

hahaha

[stable storm]

sleep good

[old carbon]

u too bro

[devout snow]

@old carbon Has your question been resolved?

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[orchid snow]

why was my integral on the LHS wrong

ik the correct one is like arc tan something

but how do I apply that in this case? !

[quaint citrus]

$\frac 1{1+4y^2} = \frac 1{1 + (2y)^2}$

[quaint citrus]

Then u can do a sub

@orchid snow

[devout snow]

@orchid snow Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

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• Show your work, and if possible, explain where you are stuck.
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[tall linden]

How is IP even NP-complete? I know that you can encode SAT as an IP program, proving its NP-hardness. But is there a proof that it cannot be harder than NP-complete problems? My thought is that verifying the problem should be co-NP complete. Given the optimal answer, you can plug it into the function to be minimized to get the maximum value. The answer is valid if there is no integral point inside the polytope given by the constraints + a new constraint that the optimization function to be larger than the maximum value. But, that's NP-complete because you can encode SAT in it.

[devout snow]

@tall linden Has your question been resolved?

[tall linden]

<@&286206848099549185>

[devout snow]

@tall linden Has your question been resolved?

[tall linden]

Maybe I'm missing something, like the information of a optimal point somehow reduces the problem into NP? I have a hunch that this is co-2QBF complete. But I have no proof.

[bright tartan]

what is IP exactly?

interactive proof?

wiki says that thats pspace, not np https://en.m.wikipedia.org/wiki/IP_(complexity)

[devout snow]

@tall linden Has your question been resolved?

[tall linden]

Integer Programming, sorry. As in maximizing a function while having integral solution that follows the constraint.

[devout snow]

@tall linden Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[safe osprey]

Currently stuck at this point, I need help understanding the last step.

[woven radish]

kisnar

[nova glen]

which step? how to get from what you have written on paper to the answer choice?

[safe osprey]

Yes

[nova glen]

you have x/x, that cancels. so far do you follow?

[safe osprey]

Yes

[nova glen]

now you factor out 2 from both the numerator and denominator

2x + 2 = 2(x+1)

4x-4 = 2(2x-2)

then you have 2/2, which also cancels